Vector Space of Real Numbers

Lessons

Senior 4 · Mathematics

Vector Space of Real Numbers

5 lessons

Lesson 1 of 5

Vectors in ℝ² and operations on them

25 min

Pulls that add up

In a tug-of-war every member of a team pulls along the same rope, so the team's pulls simply add: five people pulling with $300\,\mathrm{N}$ each give $1500\,\mathrm{N}$ . The other team pulls the opposite way, and only the difference moves the rope. Now tie two ropes to a heavy crate and pull them in different directions. The crate does not move in the direction of either rope, and the size of the result is not the sum of the two sizes. To add pulls that point in different directions, you need to add vectors.

In S2 you drew vectors as arrows and added them nose to tail. At A-level we describe a vector in the plane by an ordered pair of real numbers, define two operations on these pairs, and list the rules the operations obey. A set with two operations that obey those rules is called a vector space; it is the setting for the whole of this unit.

Definition

The set $\mathbb{R}^2$ : the set of all ordered pairs of real numbers,

\mathbb{R}^2 = \{ (x, y) : x \in \mathbb{R},\ y \in \mathbb{R} \}.

An element $\vec{u} = (u_1, u_2)$ of $\mathbb{R}^2$ is a vector; $u_{1}$ and $u_{2}$ are its components. It may also be written as a column, $\vec{u} = \begin{pmatrix} u_1 \\ u_2 \end{pmatrix}$ . Drawn from the origin $O$ , it is the arrow to the point $(u_{1}, u_{2})$ , the position vector of that point.

Equality: $(u_1, u_2) = (v_1, v_2) \iff u_1 = v_1 \text{ and } u_2 = v_2$ . The zero vector is $\vec{0} = (0, 0)$ .

Example: $\vec{u} = (3, -2)$ has components $3$ and $- 2$ ; it is the arrow from $O$ to the point $(3, - 2)$ .

Definition

Vector addition and scalar multiplication: for $\vec{u} = (u_1, u_2)$ , $\vec{v} = (v_1, v_2)$ in $\mathbb{R}^2$ and a real number $k$ (a scalar),

\vec{u} + \vec{v} = (u_1 + v_1,\ u_2 + v_2), \qquad k\vec{u} = (k u_1,\ k u_2).

The opposite of $\vec{u}$ is $-\vec{u} = (-1)\vec{u} = (-u_1, -u_2)$ , and subtraction means adding the opposite: $\vec{u} - \vec{v} = \vec{u} + (-\vec{v}) = (u_1 - v_1,\ u_2 - v_2)$ .

Example: $(3, 1) + (1, 2) = (4, 3)$ , $\ 4(1, -2) = (4, -8)$ , $\ (3, 1) - (1, 2) = (2, -1)$ .

What the operations look like

Two ropes pull a crate with forces $\vec{u} = (300, 100)$ and $\vec{v} = (100, 200)$ , in newtons, with the $x$ -axis pointing east and the $y$ -axis north. The crate feels one force, the resultant

\vec{u} + \vec{v} = (300 + 100,\ 100 + 200) = (400, 300)\ \mathrm{N}.

In the diagram one grid square stands for $100\,\mathrm{N}$ . The sum is the diagonal of the parallelogram built on $\vec{u}$ and $\vec{v}$ (the parallelogram law); equally, it is what you reach by drawing $\vec{v}$ from the tip of $\vec{u}$ (the triangle law). Adding components is the same thing written in numbers.

Multiplying by a scalar $k$ stretches the arrow by the factor $∣ k ∣$ and keeps its direction when $k \gt 0$ , reverses it when $k \lt 0$ . So every multiple $k\vec{u}$ with $\vec{u} \ne \vec{0}$ lies along the same line as $\vec{u}$ : non-zero vectors that are multiples of each other are called parallel (or collinear).

The vector from one point to another

For points $A (a_{1}, a_{2})$ and $B (b_{1}, b_{2})$ , going from $O$ to $A$ and then from $A$ to $B$ is the same journey as going from $O$ to $B$ : $\overrightarrow{OA} + \overrightarrow{AB} = \overrightarrow{OB}$ . Hence

\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (b_1 - a_1,\ b_2 - a_2),

"tip minus tail". For $A (1, 4)$ and $B (6, 1)$ , $\overrightarrow{AB} = (6 - 1,\ 1 - 4) = (5, -3)$ .

Worked example 1: combining operations

Let $\vec{u} = (2, -1)$ and $\vec{v} = (-3, 4)$ . Find (a) $\vec{u} + \vec{v}$ (b) $3\vec{u}$ (c) $2\vec{u} - 3\vec{v}$ .

Solution.

(a) $\vec{u} + \vec{v} = (2 + (-3),\ -1 + 4) = (-1, 3)$ .

(b) $3\vec{u} = (3 \times 2,\ 3 \times (-1)) = (6, -3)$ .

Worked example 2: an unknown vector and a fourth vertex

(a) Find the vector $\vec{w}$ such that $2\vec{w} + (1, -3) = (7, 5)$ .

(b) $A (- 1, 2)$ , $B (3, 3)$ and $C (4, - 1)$ are three vertices of the parallelogram $A B C D$ . Find $D$ .

Solution.

(a) Add $(- 1, 3)$ to both sides: $2\vec{w} = (7, 5) - (1, -3) = (6, 8)$ . Multiply by $\frac{1}{2}$ : $\vec{w} = (3, 4)$ . Check: $(6, 8) + (1, - 3) = (7, 5)$ . ✓

(b) In a parallelogram $A B C D$ , opposite sides are equal vectors: $\overrightarrow{AD} = \overrightarrow{BC}$ . So $D - A = C - B$ and

D = A + C - B = (-1 + 4 - 3,\ 2 + (-1) - 3) = (0, -2).

Check: $\overrightarrow{AB} = (4, 1)$ and $\overrightarrow{DC} = (4 - 0,\ -1 - (-2)) = (4, 1)$ , equal as they must be. ✓

The rules the operations obey

Because each component is an ordinary real number, the operations inherit the rules of real arithmetic. For all $\vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^2$ and all scalars $k, m \in \mathbb{R}$ :

No.	Rule	Name
A1	$\vec{u} + \vec{v} \in \mathbb{R}^2$	closure under addition
A2	$\vec{u} + \vec{v} = \vec{v} + \vec{u}$	commutativity
A3	$(\vec{u} + \vec{v}) + \vec{w} = \vec{u} + (\vec{v} + \vec{w})$	associativity
A4	$\vec{u} + \vec{0} = \vec{u}$	zero vector (identity)
A5	$\vec{u} + (-\vec{u}) = \vec{0}$	opposite (inverse)
M1	$k\vec{u} \in \mathbb{R}^2$	closure under scalar multiplication
M2	$k(\vec{u} + \vec{v}) = k\vec{u} + k\vec{v}$	distributivity over vector addition
M3	$(k + m)\vec{u} = k\vec{u} + m\vec{u}$	distributivity over scalar addition
M4	$k(m\vec{u}) = (km)\vec{u}$	associativity of scalars
M5	$1\vec{u} = \vec{u}$	unit scalar

Each rule is proved by writing out components. For instance, M3:

(k + m)\vec{u} = \big((k + m)u_1,\ (k + m)u_2\big) = (k u_1 + m u_1,\ k u_2 + m u_2) = (k u_1, k u_2) + (m u_1, m u_2) = k\vec{u} + m\vec{u},

where the middle step is the distributive law for real numbers. A1 to A5 say that $(\mathbb{R}^2, +)$ is a commutative group, which you met in the unit on binary operations.

Important

Definition (real vector space). A set $V$ with an addition $\vec{u} + \vec{v}$ and a multiplication by real scalars $k\vec{u}$ is a vector space over $\mathbb{R}$ when the ten rules A1 to A5 and M1 to M5 hold for all its elements and all real scalars.

Theorem. $\mathbb{R}^2$ with componentwise addition and scalar multiplication is a vector space over $\mathbb{R}$ .

The rules, not the arrows, are what matter. The polynomials $a + b x$ of degree at most 1 also form a vector space, and they behave exactly like the pairs $(a, b)$ : $(2 + 3 x) + (1 - x) = 3 + 2 x$ matches $(2, 3) + (1, - 1) = (3, 2)$ .

Giuseppe Peano (1858–1932), around 1910. In his 1888 book on geometrical calculus he gave the first definition of a vector space in modern form, as a set whose operations obey a list of rules. Photo: unknown author, public domain, via Wikimedia Commons.

Worked example 3: an operation that breaks a rule

On $\mathbb{R}^2$ keep the usual addition but define a new scalar multiplication $k \odot (x, y) = (kx,\ y)$ . Is $\mathbb{R}^2$ a vector space with these operations?

Solution. Test M3 with $\vec{u} = (x, y)$ :

(k + m) \odot \vec{u} = \big((k + m)x,\ y\big), \qquad k \odot \vec{u} + m \odot \vec{u} = (kx, y) + (mx, y) = \big((k + m)x,\ 2y\big).

These differ whenever $y \ne 0$ . A counter-example settles it: $k = m = 1$ , $\vec{u} = (0, 1)$ gives $2 \odot \vec{u} = (0, 1)$ but $1 \odot \vec{u} + 1 \odot \vec{u} = (0, 2)$ . So M3 fails and this is not a vector space. (M1, M2, M4 and M5 all still hold; one failed rule is enough.)

Warning

Mixing up scalars and vectors. You can add two vectors, and multiply a vector by a scalar, but " $\vec{u} + 3$ " has no meaning, and nothing in this unit divides one vector by another. To prove that a rule holds you must argue for all vectors and scalars, usually with components; to prove that it fails, one numerical counter-example is enough.

Tip

To find a missing point, translate the picture into "tip minus tail" equations: $\overrightarrow{AB} = B - A$ . Every parallelogram, midpoint or "find $D$ " question then becomes one line of component arithmetic.

Professor Dave Explains: Understanding Vector Spaces (about 9 min). Watch for the list of rules a set must satisfy to be a vector space, and for examples of vector spaces that are not arrows at all.

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Practice

Try each one first, then open the answers. They get harder as you go.

Let $\vec{u} = (4, -2)$ and $\vec{v} = (-1, 3)$ . Find (a) $\vec{u} + \vec{v}$ (b) $\vec{u} - \vec{v}$ (c) $-3\vec{v}$ (d) $2\vec{u} + 5\vec{v}$ .
For $A (1, 4)$ , $B (- 2, 0)$ and $C (5, - 3)$ , find $\overrightarrow{AB}$ , $\overrightarrow{BC}$ and $\overrightarrow{AC}$ , and verify that $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$ .
Find the real numbers $x$ and $y$ if $(2x - 1,\ y + 3) = (5,\ -2y)$ .
Find the vector $\vec{w}$ if $3\vec{w} - (2, -6) = (4, 9)$ .
Three forces $\vec{F}_1 = (2, 5)$ , $\vec{F}_2 = (-4, 1)$ and $\vec{F}_3$ , in newtons, act on a ring that stays at rest, so their sum is $\vec{0}$ . Find $\vec{F}_3$ .
$P (2, 1)$ , $Q (5, 2)$ and $R (3, 6)$ are vertices of the parallelogram $P Q R S$ . Find $S$ .
Prove, using components, that $k(\vec{u} + \vec{v}) = k\vec{u} + k\vec{v}$ for every real $k$ and all $\vec{u}, \vec{v} \in \mathbb{R}^2$ .
On $\mathbb{R}^2$ define $\vec{u} \oplus \vec{v} = (u_1 + v_2,\ u_2 + v_1)$ . Show that $\oplus$ is not commutative, and conclude that $\mathbb{R}^2$ with $\oplus$ and the usual scalar multiplication is not a vector space.

Show answersHide answers

(a) $(3, 1)$ (b) $(5, - 5)$ (c) $(3, - 9)$ (d) $(8, - 4) + (- 5, 15) = (3, 11)$ .
$\overrightarrow{AB} = (-3, -4)$ , $\overrightarrow{BC} = (7, -3)$ , $\overrightarrow{AC} = (4, -7)$ ; $(-3 + 7,\ -4 - 3) = (4, -7)$ . ✓
$2 x - 1 = 5$ gives $x = 3$ ; $y + 3 = - 2 y$ gives $y = - 1$ .
$3\vec{w} = (4, 9) + (2, -6) = (6, 3)$ , so $\vec{w} = (2, 1)$ .
$\vec{F}_3 = -(\vec{F}_1 + \vec{F}_2) = -(-2, 6) = (2, -6)\ \mathrm{N}$ .
$\overrightarrow{PS} = \overrightarrow{QR}$ , so $S = P + R - Q = (2 + 3 - 5,\ 1 + 6 - 2) = (0, 5)$ . Check: $\overrightarrow{PQ} = (3, 1) = \overrightarrow{SR}$ . ✓
$k(\vec{u} + \vec{v}) = \big(k(u_1 + v_1),\ k(u_2 + v_2)\big) = (ku_1 + kv_1,\ ku_2 + kv_2) = (ku_1, ku_2) + (kv_1, kv_2) = k\vec{u} + k\vec{v}$ , using the distributive law in $\mathbb{R}$ in each component.
Take $\vec{u} = (1, 0)$ , $\vec{v} = (0, 0)$ : $\vec{u} \oplus \vec{v} = (1, 0)$ but $\vec{v} \oplus \vec{u} = (0, 1)$ . Rule A2 fails, so this is not a vector space.

Quick check

Key points

$\vec{u} + \vec{v} = (u_1 + v_1, u_2 + v_2)$ and $k\vec{u} = (ku_1, ku_2)$ : the parallelogram (or triangle) law in numbers.
$\overrightarrow{AB} = B - A$ , "tip minus tail".
A vector space is a set whose addition and scalar multiplication obey the ten rules A1–A5 and M1–M5; $\mathbb{R}^2$ is one.
A rule is proved for all vectors with components; one counter-example shows that it fails.

Sources and credits

Written by Ganzaa: the explanations, the diagrams, the worked examples, the practice items and the questions. The unit follows the topic list of REB's S4 Core Mathematics (2026–2027); no REB text is reproduced.
History: Giuseppe Peano, J. J. O'Connor and E. F. Robertson, MacTutor History of Mathematics, University of St Andrews.
Photo: Men pulling a rope during a tug of war game, Samson Ssemakadde, Wikimedia Commons, CC0 1.0.
Photo: Giuseppe Peano, unknown author, Wikimedia Commons, public domain.
Video: Understanding Vector Spaces, Professor Dave Explains, embedded from YouTube.

1Pulls that add up

2What the operations look like

3The vector from one point to another

4The rules the operations obey

Pulls that add up

What the operations look like

The vector from one point to another

The rules the operations obey